I found a really nice way to do it linking to functional equations. Call the quartic f(x). If you notice that f(x) = x^4 f(1/x) then rearranging gives x^4 = f(x)/f(1/x). Hence the only time f(x) could be zero is when x=0. But checking this in the original equation yields f(0) = 1 so this is not possible.
I instantly recognised the coefficients [1,2,3,2,1] as being the result of squaring [1,1,1] (as it is the same process used when rolling multiple dice together, and I happen to have solved a problem related to this using polynomials) So I know that it is (x^2+x+1)^2=0, So (-1±√(-3))/2
Would appreciate if someone checked if my solution makes sense. I tried to apply factor theorem for some imaginary roots. (k is a real number) f(ki)=k^4 -3k^2+1=0 (the 2x^3 and 2x will always cancel which is nice). for the quadratic in k^2 , the discriminant = 5 so there are solutions 4 solutions for k (as there is the plus or minus in the quadratic for k^2 then plus or minus when rooting). 4 imaginary solutions so no real solutions. thank you for reading ❤
yes, thanks for spotting the mistake. ig u would need to use a+bi and then show real conpoand imaginary component were equal to 0 for some complex number
I did do it in a completely different way... First I tried to get rid of the 2x³ by noting that (x + 0.5)^4 gives x^4 + 2x³ + ... That required some calculations, but in the end, it resulted in x^4 + 2x^3 + 3x^2 + 2x + 1 = (x + 0.5)^4 + 1.5x^2 + 1.5x + 0.9375 Then I factored out the 1.5 and completed the square: = (x + 0.5)^4 + 1.5 (x^2 + x + 0.625) = (x + 0.5)^4 + 1.5 ( (x + 0.5)^2 - 0.5^2 + 0.625) = (x + 0.5)^4 + 1.5 ( (x + 0.5)^2 + 0.375) The end result obviously is always positive for all real x, so the equation can't have real solutions.
Note that 2x^3+3x^2+2x= -x^4-1. Add this equation to the original to get x^4+4x^3+6x^2+4x+1= -x^4-1. The left side factors to (x+1)^4, which is always greater than or equal to zero, and the right side is always negative, so there are no real solutions.
The symmetry of the coefficients reminded me of (x+y)^2 = x^2+2xy+y^2. So I thought it was probably a quadratic squared. I did (ax^2+bx+c)^2 = the quartic and found a,b,c=0. x^2 +x+1=0 has no solutions
A polynomial with real coefficients should factor into irreducible quadratics and linear terms. You'd start off factoring into (x^2 + ax + 1)(x^2 + bx + 1), and the a and b can be easily determined by expansion as the other comments demonstrate.
good question, I'm not sure if I have a satisfactory answer. From doing similar problems I saw the symmetry in the numbers which prompted this technqiue.
Could one utilise the AM-GM inequality by considering the product of roots and sum of roots and showing there are no real roots which satisfy the inequality
@@dharmannvids nice idea, but I don't think this works as AM-GM would require the roots to be all positive. But it's somewhat clear from the equation already that it can't have any positive real roots.
That is a nice way to solve it but there is a much simpler way. What you need to understand is that this function must cross or touch y axis. X to even power (4, 2) yields positive value. For high x values negative or positive x to the 4th power will dominate yielding a positive output for the function. The question is is there a low x value that will yield a negative result. You try 0 and you get 1, you try negative 1 x to and you get 1, you try negative 2 you get a positive result 9. You can try negative 3 as well but the coefficients of odd power arguments will not compensate for the exponents thus the function will have positive values bigger than for all real number I.e no real solutions.
@ you are right but same logic applies between 0 and -1 you want to worry about +1 which dominates, the best case scenario negative for odd exponents at - 0.5. The 3 times squared will ways be higher than 2 times third power because both exponent and coefficient so there is no chance it is lower.
@@abdullahalghamdi8062 "the best case scenario negative for odd exponents at - 0.5" Sorry, I don't understand - why? "The 3 times squared will ways be higher than 2 times third power because both exponent and coefficient" Pardon? Sorry, I don't understand what this is supposed to mean?
Let me explain it in a different way, apologies for the poor English. Between 0 and negative 0.5, +1 will always be higher than the other 2x. The rest of the terms must be higher than zero because the higher the power the lower the value and the coefficient of the term with the second power is higher than the term with third power so there is no chance the value could become negative. From negative 0.5 and negative 1, the terms with odd exponents 2x^3+2x factor out to 2x(x^2+1), if you compare this term with 3x^2+1 it will be positive because it is the same term with lower coefficient and multiplied by a fraction; therefore,it will be higher than zero adding to that the fourth power term the function out put must remain above zero.
@@abdullahalghamdi8062 "2x(x^2+1), if you compare this term with 3x^2+1 it will be positive because it is the same term with lower coefficient and multiplied by a fraction" But for x between -1 and -0.5, the 2x in front of the third expression will be between -2 and -1. So it isn't clear at once that 3x²+1 will have a bigger absolute value than 2x (x² + 1), despite the coefficient 3.
I found a really nice way to do it linking to functional equations. Call the quartic f(x). If you notice that f(x) = x^4 f(1/x) then rearranging gives x^4 = f(x)/f(1/x). Hence the only time f(x) could be zero is when x=0. But checking this in the original equation yields f(0) = 1 so this is not possible.
I instantly recognised the coefficients [1,2,3,2,1] as being the result of squaring [1,1,1] (as it is the same process used when rolling multiple dice together, and I happen to have solved a problem related to this using polynomials)
So I know that it is (x^2+x+1)^2=0,
So (-1±√(-3))/2
@@n00bxl71 oooh nice spot!
Would appreciate if someone checked if my solution makes sense. I tried to apply factor theorem for some imaginary roots. (k is a real number) f(ki)=k^4 -3k^2+1=0 (the 2x^3 and 2x will always cancel which is nice). for the quadratic in k^2 , the discriminant = 5 so there are solutions 4 solutions for k (as there is the plus or minus in the quadratic for k^2 then plus or minus when rooting). 4 imaginary solutions so no real solutions. thank you for reading ❤
@@Theproofistrivial nice solution! I hadn't thought of this!! Very slick!
How does the 2x^3 and 2x term cancel out? You have -2k^3i +2ki and you can’t add them together
yes, thanks for spotting the mistake. ig u would need to use a+bi and then show real conpoand imaginary component were equal to 0 for some complex number
Oh yes, you're right, they don't cancel. My bad! Sorry for false hope!
I did do it in a completely different way...
First I tried to get rid of the 2x³ by noting that (x + 0.5)^4 gives x^4 + 2x³ + ... That required some calculations, but in the end, it resulted in
x^4 + 2x^3 + 3x^2 + 2x + 1 = (x + 0.5)^4 + 1.5x^2 + 1.5x + 0.9375
Then I factored out the 1.5 and completed the square:
= (x + 0.5)^4 + 1.5 (x^2 + x + 0.625) = (x + 0.5)^4 + 1.5 ( (x + 0.5)^2 - 0.5^2 + 0.625)
= (x + 0.5)^4 + 1.5 ( (x + 0.5)^2 + 0.375)
The end result obviously is always positive for all real x, so the equation can't have real solutions.
@@bjornfeuerbacher5514 ah nice, like double completing the square! Good solve!!
The problem should be reduced to the quadratic equation x²+x+1=0, where (x²+x+1)² =x⁴+2x³+3x²+2x+1=0.
@@maxgoldman8903 oh nice! I never thought to try and factorise it
Note that 2x^3+3x^2+2x= -x^4-1. Add this equation to the original to get x^4+4x^3+6x^2+4x+1= -x^4-1. The left side factors to (x+1)^4, which is always greater than or equal to zero, and the right side is always negative, so there are no real solutions.
The symmetry of the coefficients reminded me of (x+y)^2 = x^2+2xy+y^2. So I thought it was probably a quadratic squared. I did (ax^2+bx+c)^2 = the quartic and found a,b,c=0. x^2 +x+1=0 has no solutions
A polynomial with real coefficients should factor into irreducible quadratics and linear terms.
You'd start off factoring into (x^2 + ax + 1)(x^2 + bx + 1), and the a and b can be easily determined by expansion as the other comments demonstrate.
if you hadnt seen a similar question before, what would prompt you to divide through by x^2 and let alpha = x +1/x in the first place?
good question, I'm not sure if I have a satisfactory answer. From doing similar problems I saw the symmetry in the numbers which prompted this technqiue.
Could one utilise the AM-GM inequality by considering the product of roots and sum of roots and showing there are no real roots which satisfy the inequality
@@dharmannvids nice idea, but I don't think this works as AM-GM would require the roots to be all positive. But it's somewhat clear from the equation already that it can't have any positive real roots.
One could alternatively notice that x^4 + 2*x^3 + 3*x^2 + 2*x + 1 = x^2*(x+1)^2 + x^2 + (x+1)^2 + 1 - 1 = (x^2+1)*((x+1)^2+1) - 1 > 1 - 1 = 0, i.e. the equation cannot have a real solution.
@@joseluishablutzelaceijas928 hmmmm... I'm not sure if that's an obvious 'spot'
at 4:08 did you rearrange correctly. nice vid btw
@@4iden.r ah whoops, good spot!!
This has been made so simple that even a GCSE student like me would understand. lovely explaining
@@YunruiHu thank you very much!
That is a nice way to solve it but there is a much simpler way. What you need to understand is that this function must cross or touch y axis. X to even power (4, 2) yields positive value. For high x values negative or positive x to the 4th power will dominate yielding a positive output for the function. The question is is there a low x value that will yield a negative result. You try 0 and you get 1, you try negative 1 x to and you get 1, you try negative 2 you get a positive result 9. You can try negative 3 as well but the coefficients of odd power arguments will not compensate for the exponents thus the function will have positive values bigger than for all real number I.e no real solutions.
Huh? You did not rule out in any way thar there could be solutions e. g. between -1 and 0.
@ you are right but same logic applies between 0 and -1 you want to worry about +1 which dominates, the best case scenario negative for odd exponents at - 0.5. The 3 times squared will ways be higher than 2 times third power because both exponent and coefficient so there is no chance it is lower.
@@abdullahalghamdi8062 "the best case scenario negative for odd exponents at - 0.5"
Sorry, I don't understand - why?
"The 3 times squared will ways be higher than 2 times third power because both exponent and coefficient"
Pardon? Sorry, I don't understand what this is supposed to mean?
Let me explain it in a different way, apologies for the poor English.
Between 0 and negative 0.5, +1 will always be higher than the other 2x. The rest of the terms must be higher than zero because the higher the power the lower the value and the coefficient of the term with the second power is higher than the term with third power so there is no chance the value could become negative. From negative 0.5 and negative 1, the terms with odd exponents 2x^3+2x factor out to 2x(x^2+1), if you compare this term with 3x^2+1 it will be positive because it is the same term with lower coefficient and multiplied by a fraction; therefore,it will be higher than zero adding to that the fourth power term the function out put must remain above zero.
@@abdullahalghamdi8062 "2x(x^2+1), if you compare this term with 3x^2+1 it will be positive because it is the same term with lower coefficient and multiplied by a fraction"
But for x between -1 and -0.5, the 2x in front of the third expression will be between -2 and -1. So it isn't clear at once that 3x²+1 will have a bigger absolute value than 2x (x² + 1), despite the coefficient 3.